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Murljashka [212]
3 years ago
8

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

-5 ≤ x < 3 and g(x) = 2(x-4)^2 for 3 ≤ x ≤ 6.
(a) If f(1) = 3, what is the value of f(-5) ?

(b) Evaluate \int_{0}^{6} g(x)\; dx .

(c) For -5 < x < 6, on what open intervals, if any, is the graph of f both increasing and concave up? Give a reason for your answer.

(d) Find the x-coordinate of each point of inflection of the graph of f. Give a reason for your answer.

(This was on the no-calculator section of the recently-released AP Calculus AB 2018 exam so I appreciate it if you tried to limit calculator usage)

Mathematics
1 answer:
4vir4ik [10]3 years ago
5 0
A) g=f' is continuous, so f is also continuous. This means if we were to integrate g, the same constant of integration would apply across its entire domain. Over 0, we have g(x)=2x. This means that


f_{0


For f to be continuous, we need the limit as x\to1^- to match f(1)=3. This means we must have


\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2


Now, over x, we have g(x)=-3, so f_{x, which means f(-5)=17.


b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,


\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10


c) f is increasing when f'=g>0, and concave upward when f''=g'>0, i.e. when g is also increasing.

We have g>0 over the intervals 0 and x>4. We can additionally see that g'>0 only on 0 and x>4.


d) Inflection points occur when f''=g'=0, and at such a point, to either side the sign of the second derivative f''=g' changes. We see this happening at x=4, for which g'=0, and to the left of x=4 we have g decreasing, then increasing along the other side.


We also have g'=0 along the interval -1, but even if we were to allow an entire interval as a "site of inflection", we can see that g'>0 to either side, so concavity would not change.
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